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1) radquad(1-x) = radquad(4-2x) - radquad(3-x)

2)rad quad(3x+4) + radquad(x-3) = radquad(x) + radquad(3x-3)

Grazie milleee!!

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√(1-x) =√(4-2x)-√(3-x)

1-x≥0 ; x≤1

4-2x≥0 ; x≤2

3-x≥0 ; x≤3

sono accettabili le soluzioni x≤1

elevo al quadrato

(1-x) = (4-2x)+(3-x) -2√[(4-2x)(3-x)]

1-x = 4-2x+3-x - 2√(12-4x-6x+2x²)]

2√(2x²-10x+12) = 6-2x

(2x²-10x+12) = (3-x)²

2x²-10x+12 = 9+x²-6x

x²-4x+3 = (x-1)(x-3) = 0

x=3 non accettabile perché >1

x=1

== 2 ==

√(3x+4)+√(x-3) = √x+√(3x-3)

3x+4≥0 ; x≥-4/3

x-3≥0 ; x≥3

x≥0

3x-3≥0 ; x≥1

risultano accettabili le soluzioni x≥3

elevo al quadrato

(3x+4)+(x-3)+2√[(3x+4)(x-3)] = x+(3x-3)+2√[x(3x-3)]

3x+4+x-3+2√(3x²-9x+4x-12) = x+3x-3+2√(3x²-3x)

2+√(3x²-5x-12) = √(3x²-3x)

di nuovo al quadrato ed isolo l’unica radice rimasta

4+3x²-5x-12+4√(3x²-5x-12) = 3x²-3x

2√(3x²-5x-12) = x+4

12x²-20x-48=x²+16+8x

11x²-28x-64=0

x = [14± √(196+704)]/11 = (14± 30)/11

x = 44/11 = 4

x = -16/11 non accettabile

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